What is the answer to 2(2^{2x})-9(2^x)+10=0 ?

The answer to 2(2^{2x})-9(2^x)+10=0 is x=(ln(5/2))/(ln(2)),x=1

Solve for x: 2(2^{2x})-9(2^x)+10=0

The solution is x=(ln(5/2))/(ln(2)),x=1

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2 (2^{2 x}) - 9 (2^{x}) + 10 = 0

x = \frac{ln ( \frac{5}{2} )}{ln ( 2 )}, x = 1

Solution steps

2 (2^{2 x}) - 9 (2^{x}) + 10 = 0

Apply exponent rules

(2^{x})^{2} \cdot 2^{1} - 9 \cdot 2^{x} + 10 = 0

Rewrite the equation with

2^{x} = u

(u)^{2} \cdot 2^{1} - 9 u + 10 = 0

Solve

u^{2} \cdot 2^{1} - 9 u + 10 = 0 : u = \frac{5}{2}, u = 2

u = \frac{5}{2}, u = 2

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = \frac{5}{2} : x = \frac{ln ( \frac{5}{2} )}{ln ( 2 )}

Solve

2^{x} = 2 : x = 1

x = \frac{ln ( \frac{5}{2} )}{ln ( 2 )}, x = 1

2^{x + 1} - 2 (2^{x - 1}) = 2^{4 x}

1^{x} = 10

1 - e^{(- 0.75 x)} = 0.4268

4 \cdot 3^{x + 1} - 9^{x} = 27

100 = 200 e^{- 0.06 x}

What is the answer to 2(2^{2x})-9(2^x)+10=0 ?
The answer to 2(2^{2x})-9(2^x)+10=0 is x=(ln(5/2))/(ln(2)),x=1
Solve for x: 2(2^{2x})-9(2^x)+10=0
The solution is x=(ln(5/2))/(ln(2)),x=1